Question: A curve in the plane is defined parametrically by the equations $x=2\cos(-t)$ and $y=5\sin(2t)$. Find the value of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{6}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{5}{2}$ (Choice B) B $5$ (Choice C) C $-5$ (Choice D) D $\dfrac{2}{5}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=2\cos(-t)$ and $y=5\sin(2t)$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}(5\sin(2t))}{\dfrac{d}{dt}(2\cos(-t))} \\\\ &=\dfrac{10\cos{(2t)}}{2\sin(-t)} \\\\ &=\dfrac{10\cos({2t})}{-2\sin(t)} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= {\dfrac{\pi}{6}}$ : $\begin{aligned} &\phantom{=}\dfrac{10\cos{\left(2\left({\dfrac{\pi}{6}}\right)\right)}}{-2\sin\left({\dfrac{\pi}{6}}\right)} \\\\ &=\dfrac{10\cos\left(\dfrac{\pi}{3}\right)}{-2\sin\left({\dfrac{\pi}{6}}\right)} \\\\ &=\dfrac{10\cdot \dfrac12}{-2\cdot \dfrac12} \\\\ &=-5 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{6}$ is $-5$.